Problem: A parabola has vertex $V = (0,0)$ and focus $F = (0,1).$  Let $P$ be a point in the first quadrant, lying on the parabola, so that $PF = 101.$  Find $P.$
Explanation: Using the vertex and focus, we can see that the equation of the directrix must be $y = -1.$

[asy]
unitsize(3 cm);

real func (real x) {
  return(x^2);
}

pair F, P, Q;

F = (0,1/4);
P = (0.8,func(0.8));
Q = (0.8,-1/4);

draw(graph(func,-1,1));
draw((-1,-1/4)--(1,-1/4),dashed);
draw(F--P--Q);

label("$y = -1$", (1,-1/4), E);
label("$y + 1$", (P + Q)/2, E);

dot("$F = (0,1)$", F, NW);
dot("$P = (x,y)$", P, E);
dot("$(x,-1)$", Q, S);
[/asy]

Let $P = (x,y)$ be a point on the parabola.  Then by definition of the parabola, $PF$ is equal to the distance from $P$ to the directrix, which is $y + 1.$  Hence,
\[\sqrt{x^2 + (y - 1)^2} = y + 1.\]Squaring, we get $x^2 + (y - 1)^2 = (y + 1)^2.$  This simplifies to $x^2 = 4y.$

We are given that $PF = 101,$ so $y + 1 = 101,$ and hence $y = 100.$  Then $x^2 = 400.$  Since the point is in the first quadrant, $x = 20.$  Hence, $P = \boxed{(20,100)}.$